S2T2

Jim Randell 8:10 am on 28 June 2020 Permalink Reply
Tags: by: Colin   

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  Jim Randell 8:11 am on 28 June 2020 Permalink | Reply

  We can use the [[ SubstitutedExpression() ]] solver from the enigma.py library to solve this puzzle.

  The following run file executes in 96ms.

  Run: [ @repl.it ]

  #! python -m enigma -rr
  
  SubstitutedExpression
  
  # regions given
  "A + F + G = 9"
  "E + F + H = 7"
  "C + I + J = 23"
  
  # remaining regions have the same value...
  "B + G + J == D + H + I"
  # ... and sum to a square
  "is_square(B + D + G + H + I + J)"
  
  # suppress verbose output
  --template=""
  

  Solution: The numbers assigned to each line are shown below:

  

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    Jim Randell 12:38 pm on 30 June 2020 Permalink | Reply

    The puzzle as originally published in The Sunday Times was worded as follows:

    Each of the ten line segments in the picture have been given a different whole number value from 0 to 9. A figure shown in a triangle equals the sum of the three values assigned to the sides of the triangle. The figures which have not been given from the other two triangles when added together to give 30. What is the value assigned to CD?

    CD being the external edge of the 23 triangle (which is the value C in my solution).

    We can adjust the run-file accordingly:

    #! python -m enigma -rr
    
    SubstitutedExpression
    
    # regions given
    "A + F + G = 9"
    "E + F + H = 7"
    "C + I + J = 23"
    
    # the other two triangles added together to give 30
    "B + D + G + H + I + J = 30"
    
    --answer="C"
    
    # suppress verbose output
    --template=""
    

    And we find that there are 24 ways to fill out the numbers, but the value on the required edge is always 9.

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  GeoffR 4:09 pm on 29 June 2020 Permalink | Reply

  
  % A Solution in MiniZinc
  include "globals.mzn";
  
  % using the vertex labels A,B,C,D and E (as original diagram in link)
  % plus F for the central point
  
  % outside edges of the pentagon
  var 0..9:AB; var 0..9:BC; var 0..9:CD; var 0..9:DE; var 0..9:EA;
  
  % internal lines to central point F
  var 0..9:AF; var 0..9:BF; var 0..9:CF; var 0..9:DF; var 0..9:EF;
  
  constraint all_different ([AB, BC, CD, DE, EA, AF, BF, CF, DF, EF]);
  
  % three given triangles are AEF, ABF and CDF
  constraint AF + EF + EA == 7;
  constraint AB + AF + BF == 9;
  constraint CF + DF + CD == 23;
  
  % the other two triangle sides (BCF and DEF) are the same total 
  % and the two add together to give a perfect square
  
  constraint DE + EF + DF == BF + CF + BC;
  
  set of int: sq = {16, 25, 36, 49, 64, 81};
  constraint (DE + EF + DF + BF + CF + BC) in sq;
  
  solve satisfy;
  
  % AB = 0; BC = 3; CD = 6; DE = 5; EA = 1;
  % AF = 2; BF = 7; CF = 8; DF = 9; EF = 4;
  % % time elapsed: 0.06 s
  % ----------
  % ==========
  
  

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  GeoffR 1:46 pm on 8 July 2020 Permalink | Reply

  My solution below is updated for the original Sunday Times description, the solution above being the solution being based on the teaser in the book Brainteasers (2002, edited by Victor Bryant).

  
  % A Solution in MiniZinc - original Sunday Times teaser
  include "globals.mzn";
   
  % using the vertex labels A,B,C,D and E (as original ST diagram in link)
  % plus F for the central point
   
  % outside edges of the pentagon
  var 0..9:AB; var 0..9:BC; var 0..9:CD; var 0..9:DE; var 0..9:EA;
   
  % internal lines to central point F
  var 0..9:AF; var 0..9:BF; var 0..9:CF; var 0..9:DF; var 0..9:EF;
   
  constraint all_different ([AB, BC, CD, DE, EA, AF, BF, CF, DF, EF]);
   
  % three given triangles are AEF, ABF and CDF
  constraint AF + EF + EA == 7;
  constraint AB + AF + BF == 9;
  constraint CF + DF + CD == 23;
  
  % for figures in triangles not given ie triangles DEF and BCF
  constraint DE + EF + DF + BF + CF + BC == 30; 
   
  solve satisfy;
  output ["CD = " ++ show(CD) ];
  
  % CD = 9
  % ----------
  % ==========
  
  

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